To find:\(\int\cfrac{sin\,2x}{(sin^4x+cos^4x)}dx\)
Formula Used:
1. sec2 x = 1 + tan2 x
\(2.\int\cfrac{1}{1+x^2}dx=tan^{-1}x+c\)
3. sin 2x = 2 sin x cos x
Rewriting the given equation,
\(\int\cfrac{2sin\,x\,cos\,x}{sin^4x+1}\)
Let y = tan x
dy = sec2 x dx
Therefore,
\(\int\cfrac{2y}{y^4+1}dy\)
Let z = y2
dz = 2y dy
\(\Rightarrow\int\cfrac{dz}{1+z^2}\)
⇒ tan-1 z + C
Since z = y2,
⇒ tan-1(y2) + C
Since y = tan x
⇒ tan-1(tan2 x) + C
Therefore,
